Difference between revisions of "2000 AIME II Problems/Problem 9"
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Alternatively, we could let <math>z = a + bi</math> and solve to get <math>z=\cos 3^\circ + i\sin 3^\circ</math> | Alternatively, we could let <math>z = a + bi</math> and solve to get <math>z=\cos 3^\circ + i\sin 3^\circ</math> | ||
| − | Using | + | Using [[De Moivre's theorem]] we have <math>z^{2000} = \cos 6000^\circ + i\sin 6000^\circ</math>, <math>6000 = 16(360) + 240</math>, so |
<math>z^{2000} = \cos 240^\circ + i\sin 240^\circ</math> | <math>z^{2000} = \cos 240^\circ + i\sin 240^\circ</math> | ||
We want <math>z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1</math> | We want <math>z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1</math> | ||
| − | Of course, we cannot have -1 as an answer on the AIME, but they asked for the smallest | + | Of course, we cannot have -1 as an answer on the AIME, but they asked for the smallest integer greater than this value, which is 0. |
== See also == | == See also == | ||
{{AIME box|year=2000|n=II|num-b=8|num-a=10}} | {{AIME box|year=2000|n=II|num-b=8|num-a=10}} | ||
Revision as of 18:07, 3 January 2008
Problem
Given that 
is a complex number such that 
, find the least integer that is greater than 
.
Solution
Note that if z is on the unit circle in the complex plane, then 
and 
We have 
and 
Alternatively, we could let 
and solve to get 
Using De Moivre's theorem we have 
, 
, so
We want 
Of course, we cannot have -1 as an answer on the AIME, but they asked for the smallest integer greater than this value, which is 0.
See also
| 2000 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
